At 25 degree C and 1 atm, a 3.00-cm^3 syringe is plugged on the end so that the

Question

At 25 degree C and 1 atm, a 3.00-cm^3 syringe is plugged on the end so that the air inside cannot escape. The gas in the syringe is compressed by rapidly depressing the plunger so that no heat is transferred to the surrounding air. The temperature inside the syringe is observed to rise by 7.3 degree C. Assume that the pressure applied to the piston is a constant 1.2 atm. (The heat capacity of air is 1.03 J/g middot K; the average molecular weight of air is 28.8 g/mol.) What is the volume change of the gas? L How much work was done on the gas? J Determine the change in internal energy of the air in the syringe. J

Explanation / Answer

a)

d V= Vfinal - Vinitial

Vinitial = 3 cm3

V1/T1 = V2/T2

V2 = V1*(T2/T1) = 3*((298+7.3)/(298) = 3.073489 cm3

so..

dV = 3.073489 - 3 = 0.073489 cm3

change to mL --> 0.073489 cm3 = 0.073489 mL --> 0.073489*10^-3 L = 7348*10^-5 L

so...

b)

W = -P*dV

W = -(1.2*101325)( 7348*10^-8) = -8.9344 J

c)

dU = Q-W

Q = 0

dU = = -8.9344 J

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